## anonymous 4 years ago Rationalize the denominator: cubic root of 2y^4/6x^4

1. anonymous

$\sqrt[3]{2y ^{4}/6x ^{4}}$

2. anonymous

My instinct is to go down instead of making it $x ^{64}$ because that would look dumb, unless it's correct but I've never had to decrease... do I just multiply by a negative exponent?

3. anonymous

=cube root (2y^4)/cube root(6x^4) =cube root(2y^4)*cuberoot(6x^4)/cube root(6x^4)*cube root(6x^4) =cube root(2y^4)*cuberoot(6x^4)/6x^4 basically separate top and bottom roots multiply by conjugates fix ugly maths

4. anonymous

cuberoot(y/3x)^4 =(y/3x)cuberoot(y/3x)=(ycuberoot3xy)/3x

5. anonymous

2/6 simplifie by 2 and get 1/3

6. anonymous

ok ?

7. anonymous

Hmm, well that is handy, didn't know oyu could simplify it but what would the correct conjugate be?

8. anonymous

you*

9. anonymous

good luck bye

10. anonymous

1/sqrt3 =(sqrt3)/3 ---like example

11. anonymous

ok ?

12. anonymous

No actually, I'm confused... can you just tell me what I need to multiply by to get rid of the radical in the denominator?

13. anonymous

1/sqrt3 so when there is radical in denominator multiplie numerator and denominator by conjugate so in this case by sqrt3 and in this way you get in denominator 3 and in numerator sqrt3

14. anonymous

because sqrt3 * sqrt3 =3

15. anonymous

I know... but what would the conjugate be. I know how to do all this already i get it I just need to know if the conjugate I am using is correct...

16. anonymous

17. anonymous

check my answer step by step

18. razor99

we have to simplify that.

19. anonymous

Ok look, the denominator is cubic root of 6x^4 .... so what do I multiply it by to take a perfect cubic root?

20. anonymous

sorry that is wrong

21. anonymous

$\sqrt[3]{36x^2}$

22. saifoo.khan

If sat is here, then there was no need of calling me! :)

23. anonymous

then you will get in the denominator a perfect cube, namely $\sqrt[3]{216x^6}$

24. anonymous

He just got here thanks anyway

25. anonymous

you got this?

26. anonymous

you have $\sqrt[3]{6x^4}$ so to make it a perfect cube, you need the exponent on the x to be 6, and you have to multiply by 6^2=36 to make the constnant a cube

27. anonymous

Wait why does the exponent for x have to be 6?

28. anonymous

because you have a cube root so you need a number divisible by 3

29. anonymous

i mean an exponent divisible by 3

30. anonymous

$\sqrt[3]{216x^6}=6x^2$

31. anonymous

So that would come out as $6x ^{2}$ ?

32. anonymous

yes

33. anonymous

Ah thank you that is ALL I needed, I guess nobody understood what I meant. Thank you very much

34. anonymous

yw